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Nuclei


Physics. 438. 13.1 INTRODUCTION. In the previous chapter, we have learnt that in every atom, the ... various properties of nuclei such as their size, mass and stability, and ... and its constituents, ∆M, is called the mass defect, and is given by ... apart a nucleus in this way, the nuclear binding energy is still a convenient.


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Given that

Mass of copper coin ^{63}_{29} C, m = 3 g

Atomic mass of copper A = 62.92960 amu

Step 1: Compute the number of atoms given grams and units of atomic mass

Number of atoms = \frac{NA}{A} * \text{ mass }

Where, NA - Avogadro's number ( 6.023 * 10^{23}) atoms/g , A - atomic mass

= \frac{6.023 * 10^{23}}{62.92960} * 3

= 2.868 * 10^{22} atoms

Step 2: Finding the mass defect of all nucleus present in the coin

Number of protons = 29

Number of neutrons = 34

Mass of proton m_H = 1.007825 u

Mass of neutron m_n = 1.008665 u

Mass defect of nitrogen \Delta m = m_H + m_n - m

\Delta m=29*(1.007825)+34*(1.008665)-62.92960

\Delta m=0.592 amu

Mass defect of all nucleus present in the coin \Delta m = 0.592 * 2.868 * 10^{22}

\Delta m = 1.698 * 10^{22} amu

Converting atomic mass units to MeV/c^2

1 amu = 931.5 MeV/c^2

\Delta m=1.698\cdot10^{22}\ \cdot931.5\ MeV/c^2

\Delta m=1.581\ \cdot\ 10^{25}\ MeV/c^2

Step 3: Determining binding energy of the given nucleus

Binding energy \Delta E_d = \Delta mc^2

\Delta E_d=1.581\ \cdot10^{25}\ MeV/c^2*c^2

\Delta E_d=1.581\cdot\ 10^{25}\ MeV

Therefore, binding energy of the nucleus of the coin \Delta E_d=1.581\cdot\ 10^{25}\ MeV\Delta E_d=1.581\ MeV

Converting MeV to joules

1 MeV = 1.6 * 10^{-13} J

\Delta E_d=1.581\cdot10^{25}\cdot\ 1.6\cdot\ 10^{-13}\ J

\Delta E_d = 2.53 * 10^{12} J

Hence, energy required to separate all the protons and neutrons is   \Delta E_d = 2.53 * 10^{12} J